Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 months ago

Probability Question?

I have a question. In letter B "There are 16 possible outcomes, and the number of combinations that have two female and two male puppies is 4C2."

What is the explanation of 4C2? Why 4C2?

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  • nbsale
    Lv 6
    4 months ago

    It's a binomial distribution: Puppies are either male or they are not.

    Let p= prob(male) = 1/2, and for clarity, write q = prob(female) = 1-p.

    For 4 puppies, the distribution is described by the terms of (p+q)^4.

    The term representing k male puppies is (4Ck) p^k q^(4-k).

    So you get 4C2 (1/2)^2 (1/2)^2 = 6/16 = 3/8

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  • 4 months ago

    Ok, so, you first need to actually understand what 4C2 means, both in terms of math and in terms of English. For the math definition, nCr = n!/(r! * (n-r)!), so in our case, 4C2 = 4!/(2! * 2!) or 6. Then, for the English definition, nCr means n choose r. That means that you are picking r items from a total of n items and nCr finds the number of distinct ways to do so.

    So in this question, you have a total of 4 puppies, so n is 4. And, you want to choose 2 of those 4 to be male, so r is 2. Then, after the males are chosen, you have 2 puppies remaining so n becomes 2. You want 2 females, so r is also 2. So, the total is technically 4C2 * 2C2, but because 2C2 is 1, it can be neglected. That is why it is 4C2.

    Hope that helped!

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  • It's a binomial distribution.

    (a + b)^n = (n! / (0! * (n - 0)!)) * a^(n - 0) * b^0 + (n! / (1! * (n - 1)!)) * a^(n - 1) * b^1 + .... + (n! / (n! * (n - n)!)) * a^(n - n) * b^n

    The probability of having either a male or a female is 0.5, no matter what.  The coefficient of each term is figured using the combinatorial.

    4C2 * (1/2)^(4 - 2) * (1/2)^(2) =>

    (4! / (2! * (4 - 2)!)) * (1/2)^2 * (1/2)^2 =>

    (24 / (2 * 2)) * (1/16) =>

    6 * (1/16) =>

    3/8

    Having all males:

    4C0 * (1/2)^(4 - 0) * (1/2)^0 =>

    1 * (1/16) =>

    1/16

    Having all females

    4C4 * (1/2)^(4 - 4) * (1/2)^4 =>

    1 * (1/16) =>

    1/16

    3 males, 1 female:

    4C1 * (1/2)^(4 - 1) * (1/2)^1 =>

    4 * (1/16) =>

    1/4

    3 females, 1 male

    4C3 * (1/2)^(4 - 3) * (1/2)^3 =>

    4 * (1/16) =>

    1/4

    Add all of the probabilities together

    3/8 + 1/4 + 1/4 + 1/16 + 1/16 =>

    3/8 + 1/2 + 1/8 =>

    4/8 + 1/2 =>

    1/2 + 1/2 =>

    1

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  • 4 months ago

    You are making Pascal's triangle for four items.

    4 choose 0 = 4!/(0!(4-0)!) = 1 There is one way none are male.

    4C1 = 4!/(1!(4-1)!) = 4!/3! = 4 There are four ways one can be male.

    4C2 = 4!/(2!(4-2)!) = 4!/(2!2!) = 6 There are six ways two can be male!

    4C3 = 4!/(3!(4-3)!) = 4!/3! = 4 There are four ways three can be male.

    4C4 = 4!/(4!(4-4)!) =4!/4! = 1 There is one way all are male.

    1+4+6+4+1 = 16 possibilities

    6/16 = 3/8 probability 2 of 4 are male

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